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%I #21 Jul 09 2019 19:28:08
%S 1,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,3,2,2,2,2,3,1,1,1,
%T 1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,3,2,2,2,2,3,1,1,1,1,2,1,1,1,1,2,1,1,
%U 1,1,2,1,1,1,1,3,2,2,2,2,3,1,1,1,1,2
%N Number of ways of writing n as a sum of powers of 5, each power being used at most five times.
%C Also known as the hyper 5-ary partition sequence, often denoted h_5(n).
%C Contains A002487 as a subsequence.
%H Timothy B. Flowers, <a href="/A277873/b277873.txt">Table of n, a(n) for n = 0..10000</a>
%H K. Courtright and J. Sellers, <a href="http://www.integers-ejcnt.org/e6/e6.Abstract.html">Arithmetic properties for hyper m-ary partition functions</a>, Integers, 4 (2004), A6.
%H Timothy B. Flowers, <a href="https://www.emis.de/journals/JIS/VOL20/Flowers/flowers3.html">Extending a Recent Result on Hyper m-ary Partition Sequences</a>, Journal of Integer Sequences, Vol. 20 (2017), #17.6.7.
%H T. B. Flowers and S. R. Lockard, <a href="http://www.integers-ejcnt.org/q10/q10.Abstract.html">Identifying an m-ary partition identity through an m-ary tree</a>, Integers, 16 (2016), A10.
%F G.f.: Product_{j >= 0} (1-x^(6*5^j))/(1-x^(5^j)).
%F G.f.: Product_{j >= 0} Sum_{k=0..5} x^(k*5^j).
%F a(0)=1; for k>0, a(5*k) = a(k)+a(k-1) and a(5*k+r) = a(k) with r=1,2,3,4.
%F G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^5). - _Ilya Gutkovskiy_, Jul 09 2019
%e a(140) = 4 because 140 = 125+5+5+5 = 125+5+5+1+1+1+1+1 = 25+25+25+25+25+5+5+5 = 25+25+25+25+25+5+5+1+1+1+1+1.
%p b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
%p add(b(n-j*5^i, i-1), j=0..min(5, n/5^i))))
%p end:
%p a:= n-> b(n, ilog[5](n)):
%p seq(a(n), n=0..120); # _Alois P. Heinz_, May 01 2018
%t n:=250; r:=3; (* To get up to n-th term, need r such that 5^r < n < 5^(r+1) *) h5 := CoefficientList[ Series[ Product[ (1 - q^(6*5^i))/(1 - q^(5^i)) , {i, 0, r}], {q, 0, n} ], q]
%Y Cf. A002487, A054390, A277872.
%K nonn
%O 0,6
%A _Timothy B. Flowers_, Nov 07 2016
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